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If $f=0$ is the unique function for which the Fourier coefficients are zero then the set $phi_1, phi_2,…$ is complete
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Proposition: If the set of functions $ phi_1, phi_2,... $ has the property that
$ f = 0 $ is the only function with that all its Fourier Coefficients are zero, that
is
$C_n= 0 $ $ forall n in mathbbN$. Then $ phi_1, phi_2,... $ is complete
Where $$ C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$
we say that the set of functions $phi_1,phi_2,... $ is complete. If $$ lim_Nrightarrow infty int_a^b Big(f-sum_n=1^NC_nphi_n Big)^2 dx = 0 $$
for every function $f$ with the property that ( $f$ is square-integrable. )$$ int_a^b f^2 dx < infty$$
I don't how any idea how to prove this.
how would you prove this? ^_^
pde fourier-analysis fourier-series
asked 3 hours ago
tnt235711
345
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up vote
0
down vote
favorite
Proposition: If the set of functions $ phi_1, phi_2,... $ has the property that
$ f = 0 $ is the only function with that all its Fourier Coefficients are zero, that
is
$C_n= 0 $ $ forall n in mathbbN$. Then $ phi_1, phi_2,... $ is complete
Where $$ C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$
we say that the set of functions $phi_1,phi_2,... $ is complete. If $$ lim_Nrightarrow infty int_a^b Big(f-sum_n=1^NC_nphi_n Big)^2 dx = 0 $$
for every function $f$ with the property that ( $f$ is square-integrable. )$$ int_a^b f^2 dx < infty$$
I don't how any idea how to prove this.
how would you prove this? ^_^
pde fourier-analysis fourier-series
asked 3 hours ago
tnt235711
345
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Proposition: If the set of functions $ phi_1, phi_2,... $ has the property that
$ f = 0 $ is the only function with that all its Fourier Coefficients are zero, that
is
$C_n= 0 $ $ forall n in mathbbN$. Then $ phi_1, phi_2,... $ is complete
Where $$ C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$
we say that the set of functions $phi_1,phi_2,... $ is complete. If $$ lim_Nrightarrow infty int_a^b Big(f-sum_n=1^NC_nphi_n Big)^2 dx = 0 $$
for every function $f$ with the property that ( $f$ is square-integrable. )$$ int_a^b f^2 dx < infty$$
I don't how any idea how to prove this.
how would you prove this? ^_^
pde fourier-analysis fourier-series
asked 3 hours ago
tnt235711
345
Proposition: If the set of functions $ phi_1, phi_2,... $ has the property that
$ f = 0 $ is the only function with that all its Fourier Coefficients are zero, that
is
$C_n= 0 $ $ forall n in mathbbN$. Then $ phi_1, phi_2,... $ is complete
Where $$ C_n = frac int_a^b fphi_n dx int_a^b phi_n^ 2 dx $$
we say that the set of functions $phi_1,phi_2,... $ is complete. If $$ lim_Nrightarrow infty int_a^b Big(f-sum_n=1^NC_nphi_n Big)^2 dx = 0 $$
for every function $f$ with the property that ( $f$ is square-integrable. )$$ int_a^b f^2 dx < infty$$
I don't how any idea how to prove this.
how would you prove this? ^_^
pde fourier-analysis fourier-series
asked 3 hours ago
tnt235711
345
asked 3 hours ago
tnt235711
345
asked 3 hours ago
tnt235711
345
asked 3 hours ago
tnt235711
345
345
add a comment |Â
add a comment |Â
1 Answer
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In your definition of completeness, in the integral it should be the absolute value squared, if you also want to deal with complex-valued functions.
Suppose $f$ is a square-integrable function with all Fourier coefficients zero. Then by your definition of completeness,
$$0=lim_Ntoinftyint_a^b|f-sum_n=1^NC_nphi_n|^2,dx=int_a^b|f|^2,dx$$
which implies that $f=0$ in $L^2(a,b)$.
It follows that the only function orthogonal to all the basis elements is the zero function.
answered 2 hours ago
uniquesolution
7,458721
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
In your definition of completeness, in the integral it should be the absolute value squared, if you also want to deal with complex-valued functions.
Suppose $f$ is a square-integrable function with all Fourier coefficients zero. Then by your definition of completeness,
$$0=lim_Ntoinftyint_a^b|f-sum_n=1^NC_nphi_n|^2,dx=int_a^b|f|^2,dx$$
which implies that $f=0$ in $L^2(a,b)$.
It follows that the only function orthogonal to all the basis elements is the zero function.
answered 2 hours ago
uniquesolution
7,458721
add a comment |Â
up vote
1
down vote
In your definition of completeness, in the integral it should be the absolute value squared, if you also want to deal with complex-valued functions.
Suppose $f$ is a square-integrable function with all Fourier coefficients zero. Then by your definition of completeness,
$$0=lim_Ntoinftyint_a^b|f-sum_n=1^NC_nphi_n|^2,dx=int_a^b|f|^2,dx$$
which implies that $f=0$ in $L^2(a,b)$.
It follows that the only function orthogonal to all the basis elements is the zero function.
answered 2 hours ago
uniquesolution
7,458721
add a comment |Â
up vote
1
down vote
up vote
1
down vote
In your definition of completeness, in the integral it should be the absolute value squared, if you also want to deal with complex-valued functions.
Suppose $f$ is a square-integrable function with all Fourier coefficients zero. Then by your definition of completeness,
$$0=lim_Ntoinftyint_a^b|f-sum_n=1^NC_nphi_n|^2,dx=int_a^b|f|^2,dx$$
which implies that $f=0$ in $L^2(a,b)$.
It follows that the only function orthogonal to all the basis elements is the zero function.
answered 2 hours ago
uniquesolution
7,458721
In your definition of completeness, in the integral it should be the absolute value squared, if you also want to deal with complex-valued functions.
Suppose $f$ is a square-integrable function with all Fourier coefficients zero. Then by your definition of completeness,
$$0=lim_Ntoinftyint_a^b|f-sum_n=1^NC_nphi_n|^2,dx=int_a^b|f|^2,dx$$
which implies that $f=0$ in $L^2(a,b)$.
It follows that the only function orthogonal to all the basis elements is the zero function.
answered 2 hours ago
uniquesolution
7,458721
answered 2 hours ago
uniquesolution
7,458721
answered 2 hours ago
uniquesolution
7,458721
answered 2 hours ago
uniquesolution
7,458721
7,458721
add a comment |Â
add a comment |Â
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